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Edmond presents a mathematical structure of free energy asymptotics, involving a resolution or singularity, the partition function and free energy asymptotic. Pre-processing steps such as normalization, Laplace and Mellin transforms, and quenching are used to make the expression more manageable. The Zeta function, theta of z, can be evaluated using the Milling Transform, and the Laplace transform is used to convert the state density function to the asymptotics. By applying the triangular inequality, integral constants are determined to obtain the partition function's upper and lower bounds.

The mathematical structure of free energy asymptotics is discussed, which requires a resolution or singularity and a foundational result of singular learning theory. The partition function is the normalizing factor for the base posterior and the free energy asymptotic is the negative log of the partition function. Pre-processing steps such as normalization, the Laplace and Mellin transforms, and quenching are used to make the expression more manageable and evaluate the zeta function. A positive constant is added to the data function, which is a product of integrals with the variables ranging from 0 to 1. The result is a x to the z, where z is a power of a, multiplied by a product of integrals indexed by the variables.

The Zeta function, theta of z, can be evaluated using the Milling Transform, which involves an integral on the control c. This integral can be evaluated using the Cauchy integral formula or residue, and the limit of the radius of the semicircle is taken as it goes to infinity. This will give the static density function. The Milling Transform is equal to zero for t greater than a, and is equal to the m minus one derivative of the expression on the numerator evaluated at the pole, giving n minus 1 copy of the logs. The resolution similarity is considered in relation to how the constant a is going to affect the Kullback-Leibler divergence.

The Laplace transform is used to convert the state density function to the asymptotics of the n. By integrating from 0 to infinity, the lower bound is the log of n to the power of n minus 1 over n to the power of lambda, while the upper bound is the sum of the absolute value of the log of theta. Applying the triangle inequality, the integral is independent of n and is equal to a constant, resulting in an upper bound of log n to the power of n minus 1 over n to the power of lambda.

An inequality involving an expression with x and y dependences was discussed. By cancelling the y integral and replacing it with its maximum, the expression became smaller. The y dependence was then pulled back into the x integral, resulting in a constant and the expression becoming equal to the Laplace transform. This was used to calculate the state density function, which is an integral from 0 to n of e to the negative theta t. The upper and lower bounds of the partition function were calculated using the same argument. This theorem states that for an integral with a number of x coordinates and y coordinates, the exponent for the y coordinates must satisfy the relationship that lambda is equal to h1 + 2k1 and must be less than the prime ones. For all n greater than 1, the upper and lower bound of the partition function is given.

Edmond is presenting a calculation involving the expansion of a modulus absolute value term using the binomial theorem and applying the triangular inequality. This gives an expression with a j from 1 to n-1, which can be cancelled out at the cost of an inequality. The Dirac delta trick is used to make the state density function explicit, resulting in an integral of a two variable function with respect to c and phi, multiplied by v and an exponential term. This helps to go from quenched deterministic form back to standard form for the asymptotic formula for the partition function.

A theorem states that constants a1 and a2 exist such that for n greater than one, a1 is less than or equal to a2 multiplied by the square root of n multiplied by x to the k y to the k prime. To obtain a lower or upper bound for this expression, the modulus of a b is less than or equal to half a squared plus b squared. By setting a equal to the square root of n multiplied by x to the k y to the k prime and b equal to c, the upper bound is obtained by replacing a negative term with the right-hand side of the inequality and the lower bound is obtained by replacing the modulus with the right-hand side of the inequality. This is then combined with the log n m to the n minus one and lambda, resulting in the final lower and upper bounds.

The goal of this talk is to discuss the mathematical structure of free energy asymptotics. This requires a resolution or singularity, plus a foundational result of singular learning theory. The notation is set up and the partition function, which is the normalizing factor for the base posterior, is discussed. The free energy asymptotic is the negative log of the partition function. Two steps of pre-processing are needed to make this expression more manageable, the first being normalization.

The Laplace and Mellin transforms are used to relate the partition function Z to the state density function v and the data function ζ. The Laplace transform relates Z to v and the Mellin transform relates v to ζ. Quenching is used to replace random variables with their averages, allowing the partition function to be expressed in a deterministic form. This allows Z to be related to v and ζ, which can then be used for numerical calculations.

A positive constant (a) is added to the data function, which is a product of integrals with the variables ranging from 0 to 1. This evaluates the zeta function of a positive function (k) with even powers and a normal crossing prior, where x is a vector going from x1 to xm. The result is a x to the z, where z is a power of a, multiplied by a product of integrals indexed by the variables.

The Zeta function, theta of z, is equal to a constant called gamma multiplied by a to the z on z plus lambda to the power of m. To evaluate the static density function, the inverse mailing transform of theta is done. This involves an integral on the control c, where c is greater than negative lambda. The integral can be evaluated using the Cauchy integral formula or residue, and the limit of the radius of the semicircle is taken as it goes to infinity. This will give the static density function.

A Cauchy integral formula is used to evaluate a transform, known as the Milling Transform, which is gamma over 2 pi i of the contour c and equal to h and z t to negative z minus 1. z plus lambda to the m dz. Depending on the sign of the term e to the z log a on t, the semi-circular contour either decays to zero or does not, in which case the integral is zero. The Milling Transform is equal to zero for t greater than a and is equal to the m minus one derivative of the expression on the numerator evaluated at the pole, giving n minus 1 copy of the logs. The resolution similarity is considered in relation to how the constant a is going to affect the Kullback-Leibler divergence.

In step three of the process, the Laplace transform is used to convert the state density function to the asymptotics of the n. The Laplace transform of the state density function is equal to a constant multiplied by e to the negative n t to the lambda minus one and a log term of t to the power of n minus one. To obtain the asymptotic of the integral, two inequalities are used. The first inequality is that the integral is greater than a log term of n minus log t to the power of n minus one when the integration region is shrunk from zero to n to zero to one. The second inequality is that the integral is less than the same log term when the integration region is extended from zero to one to zero to n.

Integrating from 0 to infinity, the lower bound is the log of n to the power of n minus 1 over n to the power of lambda, while the upper bound is the sum of the absolute value of the log of theta, where j is between 1 and n minus 1. By removing the j dependence and applying triangle inequality, the integral is independent of n and is equal to a constant. This results in an upper bound of log n to the power of n minus 1 over n to the power of lambda.

This theorem states that for an integral with a number of x coordinates and y coordinates, the partition function is bounded by constants a1 and a2. The exponent for the x coordinates is h, and for the y coordinates is k'. The exponent for the y coordinates must satisfy the relationship that lambda is equal to h1 + 2k1 and must be less than the prime ones. The theorem states that for all n greater than 1, the upper and lower bound of the partition function is given.

The speaker discussed an inequality involving an expression with x and y dependences. By cancelling the y integral and replacing it with its maximum, the expression became smaller. The speaker then pulled the y dependence back into the x integral, which resulted in a constant, and the expression became equal to the Laplace transform. This was then used to calculate the state density function, which is an integral from 0 to n of e to the negative theta t, and the upper and lower bounds were calculated using the same argument as before.

The Dirac delta trick is used to make the state density function explicit. The x integral is highlighted in green and the y integral is treated as a constant. The integral of the expression is restricted to t between 0 and n radical k prime. The integration is extended beyond that region to 0 to infinity with a modulus sign. This results in a2 prime n to the lambda dt. The term is collected together with h two lambda k prime.

Edmond is presenting a calculation that involves expanding a modulus absolute value term using the binomial theorem and applying the triangular inequality to remove the dependency on the log n term of the exponent. This gives an expression with a j from 1 to n-1, which can be cancelled out at the cost of an inequality. This expression is integrable because the exponent for each of the y variables must be greater than negative one, which is the irrelevancy of the y-coordinates.

Partition function is defined as an integral of a two variable function, x and y, with respect to c and phi, two differentiable functions. The function is multiplied by v, a factor greater than zero, and an exponential term. This helps to go from quenched deterministic form back to standard form for the asymptotic formula for the partition function.

A theorem states that there exist constants a1 and a2 such that for n greater than one, a1 is less than or equal to a2 multiplied by the square root of n multiplied by x to the k y to the k prime. To obtain a lower or upper bound for this expression, the modulus of a b is less than or equal to half a squared plus b squared. By setting a equal to the square root of n multiplied by x to the k y to the k prime and b equal to c, the modulus is less than or equal to half times n to the 2k y to the 2 k prime multiplied by c squared. For the upper bound, this expression is replaced with a2 multiplied by the square root of n multiplied by x to the k y to the k prime.

In the proof of a theorem, the upper bound is obtained by replacing a negative term with the right-hand side of the inequality and absorbing the k-term multiplied by beta. The lower bound is obtained by replacing the modulus with the right-hand side of the inequality, resulting in 3 half n and e to the negative half beta c square. This is then combined with the log n m to the n minus one and lambda, giving the final result.

uh all right so uh welcome everyone to um this talk continuing from last time where we introduced the idea of the free energy and work through an example of um free energy asymptotics um that's that's the bottom row there let me that's the potential there so we were there and that's that last um layer level update is um on our goal today and then uh so that's kind of the the main that's the sort of mathematical structure for the free energy asymptotics um to to get to the actual free energy asymptotics there is some um so resolution or singularity would be needed plus um main familiar one which is uh which is a another foundational result of singular learning theory that we shall get to um eventually okay so that's our goal um so kind of a word of warning for uh the next hour it's going to be very calculation heavy um so but i i hope to emphasize the chain of ideas that lead to the result so um i guess you know audience can look either look out for those ideas and i will try to emphasize them or you can try following details and do stop me if um you know if it becomes uh unintelligible bring it on i say sorry i said bring it on okay cool um all right so let's let's let's again set up the notation and sort of review what where we were um last time um just very quickly um because this is part of the chain of ideas and it's kind of a core part of the innovation of slt so it's very very repeating so we have the true distribution and from the true distribution we uh independently sample a bunch of data n of them and then we want to model those data with um with a model parameterized by w which lives in a parameter space capital w which is a subset of the d-dimensional euclidean space and we endow that parameter space with a prior okay so um the the quantity that we want to estimate is the partition function which is the normalizing factor for the base posterior and that's the formula and where ln is the log loss and it's given by the average of the logs of the likelihood function well negative um of that so it's average of the over the empirical data okay so this is ultimately what we want to get to but that's a very unwieldy expression also um we uh the the free energy um because this is called the free energy asymptotic is is the negative log of that negative log of the partition function so um the the idea is to do two steps of pre-processing um to get to something we can handle the first step is normalization normalized which gives us normalized versions of these quantities

where the log loss becomes is normalized by the same quantity at the true distribution so w naught um so w naught is in um called the the parameter set of true distribution so it's where um p of x w naught is equal to the truth um if you if we don't want to assume realizability uh it's perfectly okay to replace this with the minima so w naught is um is in uh i'm going to abuse sensation and say that it's in the set of admin of px w okay sorry for that just to quickly write down um okay and after normalization our partition function um becomes in the realizable case it becomes this expression where kn is um so this is defined as kn in the realizable case okay um as i recently found out this normalization is also absolutely crucial to do numerical calculation um because this part here really brings the range of the order magnitude of the quantities we want to compute down to manageable level uh in numerical simulation okay so the second step of um uh pre-processing is i guess quenching invoking a term from glass manufacturing or from statistical physics um it's basically replacing all the random quantity here so uh kn uh depends on kn and lns and anything with substrate ends depends on the set of um random samples which which means that they themselves are random variables and quenching just means that replace those random variables done with the average and that gives us so kn becomes the k function the log likelihood ratio average look like a ratio function um which is that um and we get the the laplacian integral that we want so the z becomes this determinant deterministic form and that we started here well we started here last week and we are going to continue here and the idea is that once we get a handle on this we will go back up the chain and we will will be partially up the this chain of um processing okay let's go to the next board the next idea is the core observation that the conservation that allows us to get a handle on zn which is which is that last expression um is to relate it to two other quantities which are the state density density function and the data function these are the expression where delta is the direct delta and that's the zeta function the the main observation is that these quantities are related by two integral transforms so the first integral transform relating z and v is the laplace transform so um and going from v to that we use the mailing transform and um z to v the the inverse of the main transform um some caveat is that

some sort of distribution some sort this distribution theory is suppressed to make uh these connection uh precise um well um the direct delta is a such stress distribution um we are not going to talk about that um today we are just going to assume that these are established facts the idea is that if k of w is normal crossing that of the zeta function is easy to evaluate last time we evaluate the case where k of x 1 x 2 just a two-dimensional model is x1 squared x2 square uh with the prior being a constant one and today we are going to opt empty and um go to level one of the ladder uh yeah i'm going to go to the fresh new bullet to start the next level okay so level one going beyond last week is we add more variable so lastly we did a specific case of two variables where today we are going to do a case of arbitrary number of variables where x is now a vectors a vector going from x 1 up to x m and since our functions are our k function is a positive function their powers are all pos or even powers so the k's are all positive integers non-negative integers and we also generalize our prior to uh prior in normal crossing form as well so um so x1 h1 multiplied to xm h to m and we'll usually abbreviate this to x to the 2k using multi-index notation x to h and so i'm going to stick to this notation um quite consistently so x will always mean um the um this um okay let's talk about x later but n will always mean mean the uh the number of x's right okay so so let's let's do um let's go through the same uh steps we did to derive the asymptotics using the example last week but today it's going to be a much more involved computation so we start with this data function so step one evaluate this data function okay so zeta of that is equal to well it's equal to kx to the z x dx which is equal to and apology i actually wanted to write an a up the front here where a is a positive constant i i tried to drop that and motivated this later on but it's just going to make that expositionally unwieldy so i'm just going to without justification at a there and trust me it's going to be important later um okay so um that's become um a x to the 2 k o to the power of z and x to the h um dx and remembering that these are sorry these are all integral in the square or in the cube cubic region and these integrals all factorize um this becomes the a to the z just comes up the front so it becomes x to the z and it just becomes a product of integral index by the variables and they are all zero to one integral

x i to the k i z plus h i um and those are very easy integral you integrate them to get 2 k i said plus h i plus 1. okay we simplify that not simplify that to write it in a suggestive form which is characteristic of this level we write that as z plus h i plus 1 over 2k i so we factorize all the 2ks 2ki up the front and the idea is that we want this to be a very simple thing and it's in fact constant for all i so um these are equal to the same constant which we call lambda and that will in fact be our rlct when we talk about resolution so all the coordinates are in all coordinate in all the x-coordinates um the the exponents have this relationship and that's going to be that that's going to be the the meaning of x um throughout this talk okay and in that case we get a simplified expression so theta of z is equal to well some constant called gamma um which just contain all these two kis um multiplied together um multiplied by a to the z on z plus lambda to the power of m okay so that's our zeta function let's move back to the first port to to continue with step two so step two we do the inverse manual transform okay so um just to copy from the previous spot we want to inverse mailing transform this function okay it's um note that this is uh meromorphic on c in fact it only has one pole so um and lambda is a rational number the pole is at negative lambda lambda where lambda is a rational number it's a positive rational number um and that's an m order poll so now we know that we want to evaluate the stat density function by doing the inverse mailing transform of theta and so i'm going to quickly actually actually do the mailing transform the inverse magnetic transform and if we write out the the formula for investment transform oops we get this integral so we want to evaluate this integral on the control this is the control integral on the control c where the control is this control so that's the origin zero and that's c where c is basically anything um c is greater than negative lambda so anything there is is fine in fact i should just put that on the on on the on the y-axis on the imaginary axis but that's the control that we want to evaluate um and we are going to evaluate this um by um uh by by doing uh by doing that so we are going to evaluate uh this close control in which case we can use either the cauchy integral formula or residual and take the limit as this as us this radius of the radius of this semicircle goes to infinity as r goes to infinity and then that will give us our

inverse million transform uh let's let's write that out we get gamma over 2 pi i um of the contour c and it's equal to h and z t to negative z minus 1. z plus lambda to the m dz and uh this here is equal to 1 over t times e to the z log a on t okay so one of t doesn't matter so this that is the well that is the the numerator in the cauchy integral formula um for instance and the one thing to note is that um this control here if you want to evaluate this control um and claim that this circular the semi-circular contour is equal to the um uh to integrate on that straight control the diameter contour we need to prove that the the arc um contribution to the control goes to zero as r goes to infinity we can either use jordan's lemma for that or just uh or straight up just use um integral bound on that it's not hard except we do need to be careful that log t this this term here that the sign of that term um if if the sign of that term is positive so if basically saying that if t is less than a um if uh if the sign of that term is positive then that semicircular does decay to zero otherwise if it is if t is greater than a then that contribution as i've drawn the the green arc that does not decay in which case we need to close um close the contour on the other side in which case we don't pick up any residue so um so the the so v of t um is equal to zero for uh t greater than a okay the the the integral is um just zero all right um for um for for the control contribution on the other side it's just equal to either use the residue formula or because the cauchy integral formula to get that so it's the m minus one derivative of the expression on the numerator evaluated at the pole um which easily gave us um e to the lambda minus 1 a to the lambda and then because because of the n minus 1 derivative we get n minus 1 copy of the logs and this is only for 0 less than t less than a okay that's that's our milling transform okay let's go to the next board and ask a question about the a so a is constant but i guess eventually it's just a positive function of x right so i'm a bit i'm just curious i hadn't thought through this the condition where t is greater than a the density of states is just zero um um so uh um you are you are you are thinking through the resolution similarity and how how how that a is going to so um remember that k um the uh our kl divergence is going to be just straight up uh normal crossing no no vector up the front um those uh those factors out the front is

pushed to the uh prior and uh so so that will be handled in on the next level so this a is actually there for a different reason which which you will see um uh very soon okay cool yeah cool um okay next fold okay now that we got our state density function um what we will do in step three is to do the laplace transform to to get from status intensity function to uh the asymptotics of that n okay and this is going to be illustrative of the process that we will use in the next level as well so let's have a go so zn is the laplace transform of the stat density function which is equal to um i'm pulling uh the constant up the front and and this is where the a uh appears right so um it's it's zero outside of when t is greater than a so the integral only goes from zero to a and um e to the e to the negative n t t to lambda minus one and there's a a to the lambda term which does not depends on t let's put that up the front and log a on t to the power of n minus one okay so the next step is to do a substitution of t prime equal to n over a t and i'm going to after substitution i'm going to write t prime as t again um uh so that becomes um so i'm going to lump all the constant into one single constant code c um and okay let me let me write that out so we get 0 to n the terminals change from a to n because of the substitution um dt e to the negative 80 because t prime t is equal to a on n t um and we get t to the lambda minus one uh well the uh uh the constant there gets the a constant get uh pull into the constant term and uh the n to the lambda term uh get pull up the front and then this log term becomes log n over t n minus 1. okay all right so um we want the asymptotic of this integral um we are going to do it uh in a slightly different way from last time which generalized better we are going to do this as our two inequalities we are going to prove that that n is bounded by um above and below by some expression involving the n asymptotic uh expansion that we want right so then n is greater than um okay what did i do here okay it's greater than since everything in the integrand uh this is positive that is positive because t is greater than zero and that is positive as well because uh t is less than n um throughout the integral integration region so um this is certainly greater than if we just just shrink the integration region from zero to one to zero to one and um yep uh so let's just let me write that out uh log n minus log t n minus one and within this um 0 to 1

region log t is is a log t is a negative value because t is less than 1 and negative log t is a positive value and therefore this whole thing will be will be larger this whole thing will be smaller if we just remove this term right so once we remove that term we can pull log n up the front and we get the and the integrand and the integral has no dependence on n anymore so it just integrates a constant and let's lump that into a constant we call c prime um and log n to the n minus 1 get pull out the front and and lambda is still out the front okay so that's the uh what is it the low bound and for the upper bound we we do um okay so for the upper bound we we change the integral dependence on the terminal dependence on n to just by integrating it to infinity the problem is that once t is once t is larger than n that term becomes uh negative so and and what depends on the sign of on the parity of n minus 1 that that will mess with the sign of our integral a little bit but no matter we just slap on absolute value on it and that will retain actually this is just kind of a triangle inequality as well you can slap on triangle equality on the whole thing to to get the following expression so that's less than integral from 0 to infinity of that's positive that's fine of the modulus of t to the n minus 1. which is equal to and i'm going to use the binomial theorem to expand that i think there might be a quicker way in this case but that's uh i'm going to just use that use the binomial theorem so j equal to 1 to m of n minus 1 which is j of well the log n term is always positive um so and oh so uh sorry that is not equality it's an inequality again by just taking the taking the absolute value using triangular equality into the sum of the absolute value of the log theta um okay so sorry that's n minus one minus j and that's whoops and that's j okay um well this is the only dependence on uh on n inside the integral and it would be larger if we don't have this j which range which is always less than n minus one if we don't have that j um making the power uh smaller and smaller so if we just remove that and we can slap on uh inequality um and pull out log n to the n minus one 1 all the way up the front and the integral are all um independent of n and those are just integrable functions which you can check um by some integral tests okay so that's and lumped the the integration into a constant cos c prime prime and we get log n to the n minus 1 over n to the lambda okay and that that is our

asymptotic well it's the upper upper and lower bound with the same uh n dependence hence the asymptotic for z n is log n to the n minus one over n to lambda okay um cool okay let's let's go to the next level that's um i think we might be faster now that the complicated ideas are out of the way so level two i'm calling last week level zero by the way so this is level two level 2 add irrelevant variable so we are going to add a variable that we shall henceforth call y um which has exponent k primes and the prior exponent instead of h is h prime and the idea is that adding these does not change the asymptotic so recall before that the non-prime exponent satisfy the relationship lambda is equal to h1 plus 2k1 equal to the dot equal to the hm ones and okay by the way so x y um i'm going to say that use u u equal to h y and play u is the blow up coordinates because this is this is eventually um we got we we get this coordinate eventually by blowing up um these leave in zero to one to the m cross with zero to one to the s and s will henceforth be the dimension of the y coordinates did i use extra that's wrong that's y to the h prime okay um and the for the prime coordinates we shall have lambda is less than the prime ones the the corresponding expression for the prime ones okay i'm just going to use j um where j is equal to one up to s all right um so the theorem that we shall prove in this level is theorem 4.7 in the gray book and it says that um the the partition from the well the laplace integral in this case is this integral well the dx integral and then the d y integral over the s number of x's um e to the negative n y to the 2 to the k prime x to the h y to the h prime um okay this this beta is added in at the moment because we can um and because it doesn't mess with any of the calculation that we want to do but i guess more concretely and less flippantly um this is added to make the connection to thermodynamics a bit more a lot more precise and and we shall discover it's used in the future um but for now we're just adding to strengthen our result um so that's the partition function and the theorem is that there exists constants a 1 a 2 strictly positive such that for all n greater than 1 we have the following upper and lower bound okay i probably should name this um and dependency that's okay um okay let's go to the next board for the proof for because we do need some huge area for calculation okay um proof so zn that expression that is is equal to e to the negative

and beta is always positive x to the 2 k y to the 2 k prime x to the h y to the h prime well um since the integration the the y integral uh is r0 between 0 and 1 so y's are all between 0 and 1 so if we just cancel that and replace that by the by its maximum the whole thing becomes uh larger sorry the whole thing becomes smaller hence we have the inequality that the original expression is larger than uh the expression with y in the exponent replaced with one um and that's um so now that this this exponent exponential doesn't have y dependence we can pull it into the y output pull it back into the x integral and we get so collect the y integrals the o the y dependence d y y to the h prime and the x integrals here and well that's just a that's just a constant let's call it a one prime for now um and this term here is um well that's the uh that's equal to the hmm that's not exactly the laplace transform but okay that's equal to this expression here right um yeah okay i i i thought that was straight up the laplace transform but there's a step involved where um uh where we do a change of variable uh in t with two to two t over t t divided by n and that removes the n dependency um in the in the exponent there uh but but just just using the uh the definition of the direct delta distribution we get this equal equality regardless that's kind of the proof that we did last week as well so i hope that's um uh you allow me to just use that expression um i'm sold okay cool um uh right and and this is v of t if you remember and that vrt is what the is the is the state density function uh that we use uh in our previous level uh because there is only x dependence in here so we actually do know the expression for that and it is um so this is now uh equal to well the a1 prime expression is there and remember that the v of t is only um is its integral is zero beyond uh when t is larger than a where a now is the n there so it's an integral from zero to n of dt e to the negative theta t and then we just substitute in uh the expression that we found last time with a with a equal to n is t to the lambda minus 1 n to the lambda log n on t n minus 1. okay um right and we kind of repeat the argument we use to get the upper and lower bound um for us last time um so for the lower bound um sorry this is this is a prime prime now because um i i have collect collected the the constants in the density function up the front as well okay so this is equal to um a prime prime pulling the n to lambda up the front and

doing the same trick with um shrinking the integration region removing the dependence on n into 0 to 1 and and just removing removing t there and introducing uh inequality we get e to the beta t t to the lambda minus one and log n n minus one so this is just so this is equal to a prime prime a1 prime prime integral 0 to 1 dt e to the theta t t to the lambda minus 1 multiplied by log n m minus 1 and 2 lambda that whole thing there we shall call a1 which is the constant that we needed to find in our theorem okay let's move back to the first board and complete the upper bound okay so for the upper bound we know that zn can be written as so again using the dirac delta trick again to make explicit the state density function so the irrelevant coordinate is collected in the y integral and the relevant ones the x ones are collected in this so i'm going to highlight this in green okay so the idea is that this integral this last x integral well it's x integral it's not a y integral so in this integral y is treated as a constant and this is treated ours as our a in our previous level um so that whole thing becomes uh so this whole thing is let me use the same color that whole thing is equal to some constant let's call it a2 prime times the expression that we found before with a is equal to n y to the 2k so we get in our denominator y to the 2k prime to the lambda so don't forget that this is just a number because it's a normal crossing and one's value for y is given is just a number multiplied by log so a is again y 2 to 2 k prime divided by t to the n minus 1. okay and and again this integration uh that this expression is restricted to t between 0 and the a constant in which in this case again it's n radical k prime so for the upper bound we are going to do the same trick again um by saying that we are going to extend this expression beyond this um this this domain and we are going to integrate from 0 to infinity but we need to slap on modulus sign so this becomes modulus sine because beyond that region that log becomes negative also alternatively the justification is uh triangle inequality okay so this whole thing becomes uh a2 prime n to the lambda dt this is zero to infinity um so that's that originally depends on uh on n y to the 2k and we just extend it to e to the to infinity beta t the d y integral um so we are collect so this term um get collected together with uh the term up here to get h two lambda k prime um okay i'm running out of space um i'm going to write the next line lambda t to

lambda minus 1 times the the modulus absolute value term we mentioned before and and then we are going to do the same trick with um expanding this using the binomial theorem and removing the dependency on the log n of the log n term uh on the exponent okay um yeah and and taking this using triangular inequality again to take the absolute value into the binomial expansion so that's that will give us j equal to 1 to n minus 1 n minus 1 choose j [Music] oh yeah i guess his internet had an issue just rage rage quit in the middle of a calculation yeah no this is a nice calculation i saw a bit of this in the book but um uh yeah it's good to see it say it again i guess yeah it's uh edmond is presenting in a way that's a bit clearer than what's in the book i think so yeah i think i think useful yeah it looks kind of intimidating at first but it's not so bad yeah actually um when i was reading through there's not like as long as you can integrate you're good i think right yeah yeah yeah i think to people who are more well versed in this kind of thing it probably looks very routine but there's just a lot of uh some of the little integrals that are involved here are not things i'm used to doing i guess i don't spend a lot of time doing melon transforms or that kind of right right yeah i got it's kind of it would have been it would have been more funny if you'd been frustrated and in the middle of a you know difficult time it just disappeared um [Music] well i didn't work this out oh beforehand so um i shouldn't be like it's already over so um okay let me let me complete this so you want to re-attach just reattach the speaker um so that's m minus one minus j sorry uh you probably want to reattach a speaker huh my connection is not very stable uh i said uh you probably want to reattach a speaker oh okay right [Music] yep ah cool okay i should be attached uh can people hear me yep yep okay cool um okay um so cool uh i might be lagging so i'll just write um t to the lambda minus one lambda and something okay so at the cost of an inequality the the j is cancelled here and then it can be pulled up the front and this whole thing um is integrable and it is integrable because um because this y integral is integrable but that's only because um that exponent there for for each of the uh i that the y variables satisfied um to lambda k prime is greater than negative one otherwise that's not going to be integrable and that's precisely the the irrelevancy of the y-coordinates and that's

um that's that inequality that we have imposed at the beginning and and that gives us the upper bound so if you collect the whole integral into the constant we get that and and that's uh that proves proof our next level um okay so i i the the the time is up uh there's still another level that i've prepared um uh do we want to continue or do do we want to leave that for next i think we time continue okay so okay let me quickly do the next uh level then um okay level uh i think we don't we don't probably don't have apples um i guess your audio is a bit sorry uh i think we don't have a clear idea what we want to do in the working session anyway right so we might as well just keep rolling with whatever you have to present okay right uh let me actually physically move to a place where i know my wi-fi is probably better okay let's see if this place is bad all right uh level three um so what we're going to do next is to um actually make so actually go back up the the the the quenching and normalization chain um so instead of keep saying the laplace integral uh that's that's not how you spell that we are going to say that we are going to look for the asymptotic formula for the partition function but in standard form so in normal costing form basically so this is one step below actually using resolution singularity to um to obtain um uh that's in the phone okay so uh for as a definition um we have i'm going to define what um the standard format position function is so we have our let me actually use u here is equal to x y again it's uh uh expectation into um the relevant part and the irrelevant part um and we have that's the variables we have two other function which one is c and one is phi and those are both differentiable function at least once one time differentiable and v because it's going to be kind of the factor of the front for our prior is going to be greater than zero uh on the uh on the domain all right and the partition function is defined as um so it depends on those two functions um is equal to um okay so for now whenever i write the x integral it's it's in it's from zero to one uh it's in this and similarly for uh y um so that's equal to v of x y x to the h so those number crossing terms are still there e to the negative n beta so that's still the same old thing except that there is that term which is near and oops the prime coordinates are for y's and the other term which is the term that will help us um go back from the quenched deterministic stuff back into

our stochastic stuff um i'm we are going to introduce the soup gnome for uh for that because they will show up in the um okay so the theorem is uh this is theorem 4.8 in the gray book um is that okay that same form of the same form of setup there exists constant a1 a2 such that for n greater than one um we have okay a1 i'm going to okay let me not be it's not going to save me too much time um but modified by some constants are less than equal to a2 same thing it's a positive half beta um and the suit norm of that as well okay um okay uh all right that's uh okay last let's go to the next ball and it's the last chunk of calculation okay proof um so um so we want to have uh inequality for uh for this expression um let me write it in a convenient way okay square root n beta x to the k y to the k prime and then i'm going to put this here um so we we want an upper and or or lower bound for this expression so this this is easy to handle so if you want the upper bound take the uh take the sup uh if you want the lower bound take the in [Music] this part is harder to handle because ultimately we want asymptotics in n and this this n here sort of messed with the asymptotics and this part is the familiar that n part that we have done in previous level the the trick here is to um is to massage this into a term that can be absorbed into where the end dependency can be absorbed into the previous familiar term and just to observe uh that we can either use cauchy shorts but i i thought this is easier so the square of anything is of any real number is positive um uh regardless of whether or not that's plus or minus and that's equal to a squared plus minus 2 a b plus b squared hence we get that plus minus 2 a b is always less than or equal to a squared plus b squared so regardless of the sign that's that's less than the same thing that means that a b the modulus of a b is less than or equal to half a squared plus b squared and if we just set a equal to uh root n x to the k y to the k prime so notice that the case there's no two k's there because some squares have been taken and b is equal to c and we would get uh we get square root of n uh x to the k y to the k prime c um modulus is less than equal to half times n to the 2k y to the 2 k prime so the square and the square root that the square is reintroduced the square root of n is removed and um c squared so we are going to put this in there um uh for the uh for the upper bound we are just going to uh replace this whole thing with

actually for the lower bound we are going to replace this whole thing with with um with negative the modulus and then replace it with the the inequality on the right hand side okay let me just write it okay so for the upper bound we uh we have um uh this is um uh this whole thing will become larger if we replace this thing which can be positive or negative by just by just a positive part okay and um and then it will become even larger if we replace by the right hand side of the inequality and the half the half n x to x to two k y just to the k term multiplied by beta can be absorbed in here and and we get uh and we get z to the n on two because um we got negative n plus positive um half and that um and that's that's how we get uh e to the half beta and i've sneakily removed the dependency of c on x y by just taking the soup um and again the v is easily handled by taking the subname as well so similarly for the lower bound we have that term there um plus root beta and x to the k y to the k prime is greater than minus the modulus the modulus there which is greater than if we replace it by um the inequality the right-hand side of the inequality so that thing becomes um that thing looks like 3 half n and and then we add that and then and then that gives us the lower bound okay uh okay let me just write out so we get three half and um e to the negative theta on two mod c squared so there's no negative thing okay and the log n m to the n minus one and lambda are uh vary in in those in those terms and i'm going to claim that it's the end of the proof okay okay sorry sorry that's a bit rushed and no no that was great i think that's it i i had not understood that argument before yeah terrific let me just go back and look at the statement theorem and see if i so those uh okay so a1 and a2 just come from the original expression for the z right yeah okay actually uh let me let me i actually did write this all out let me let me just actually write it out then on the next show okay so we get uh what is this tree half and e to the negative half beta c square uh the minimum of v less than or equal to z and c b less than or equal to that this is a half and instead e to the half theta square okay so um this thing is is um is greater than so we want a lower bound for that once more it's the one prime so there's a constant before in our previous level and substitute n by um three half n three half and lambda which is greater than equal to well you can um the the three half uh the tree half to the lambda is